£¨6·Ö£©Ä³¿ÎÍâС×éͬѧÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨Ä³Ð©×°ÖÃÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬Ì½¾¿Ä³¿×ȸʯ[»¯Ñ§Ê½ÎªaCuCO3¡¤bCu(OH)2 £¬a¡¢bΪÕýÕûÊý]Öи÷ÔªËصÄÖÊÁ¿¹Øϵ¡£

¢Ù ΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ£»£¨×¢£º¼îʯ»Ò¿É¿´×÷ÉÕ¼îºÍÉúʯ»ÒµÄ»ìºÏÎ

¢Ú Ïò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ__¡¡¡¡¡¡¡¡¡¡________                        £»

    ±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆΪ________        £¬

ʵÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷____________________________________________£»

    ¢ÛÁ¬ÐøÁ½´Î¼ÓÈÈ¡¢¹ÄÆø¡¢ÀäÈ´¡¢³ÆÁ¿¼××°ÖõÄÖÊÁ¿£¬ÖÊÁ¿²î²»³¬¹ý0.1g£¬Åжϸÿ×ȸʯÒÑÍêÈ«·Ö½â¡£

       ¢Ü ¸ü¾«È·µÄ²â¶¨µÃ³öÈçÏÂÊý¾Ý£º¿×ȸʯÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ16.52 g±äΪ12.00 g£¬×°ÖÃÒÒÔöÖØ1£®44 g¡£Ð´³ö¸Ã¿×ȸʯµÄ»¯Ñ§Ê½________________________¡£

 

¢Ù»ò

£¨¶à¼ÓNaOHÈÜҺϴÆøÆ¿¿Û1·Ö£¬¹²1·Ö£©

       ¢Ú ½«X·Ö½â²úÉúµÄË®ÕôÆøËÍÈëÊ¢ÓÐŨÁòËáµÄÏ´ÆøÆ¿ÖУ¨1·Ö£©£»ÎÞË®ÁòËáÍ­£¨1·Ö£©£»

       X·Ö½â²úÉúµÄË®ÕôÆøÈ«²¿±»Å¨ÁòËáÎüÊÕ£¨1·Ö£©

       7CuCO3¡¤8Cu £¨OH£©2 »òCu15£¨OH£©16£¨CO3£©7»ò15CuO¡¤7CO2 ¡¤8H2O £¨2·Ö£©

½âÎö:ÂÔ

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÒÑÖªXÊÇÒ»ÖÖÑΣ¬HÊÇÒ»ÖֹŴúÀͶ¯ÈËÃñ¾ÍÒÑÕÆÎÕÒ±Á¶¼¼ÊõµÄ½ðÊôµ¥ÖÊ£¬F¡¢P¡¢JÊdz£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬I¡¢E¡¢G¶¼Êǹ¤ÒµÉÏÖØÒªµÄ¼îÐÔÎïÖÊ£¬PºÍJÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉI£®ËüÃÇÓÐÈçϵĹØϵ£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©GµÄ»¯Ñ§Ê½Îª
NaOH
NaOH
£»
£¨2£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù
2Na2O2+2CO2=2Na2CO3+O2
2Na2O2+2CO2=2Na2CO3+O2
£»
¢Ú
2NH3+3CuO=N2+3Cu+3H2O
2NH3+3CuO=N2+3Cu+3H2O
£»
£¨3£©Ä³¿ÎÍâС×éͬѧÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨Ä³Ð©×°ÖÃÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬½øÒ»²½Ì½¾¿XÖи÷ÔªËصÄÖÊÁ¿¹Øϵ£®
¢ÙΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ£»
¢ÚÏò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ
½«X·Ö½â²úÉúµÄË®ÕôÆøËÍÈëÊ¢ÓÐŨÁòËáµÄÏ´ÆøÆ¿ÖÐ
½«X·Ö½â²úÉúµÄË®ÕôÆøËÍÈëÊ¢ÓÐŨÁòËáµÄÏ´ÆøÆ¿ÖÐ
£»±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆΪ
ÎÞË®ÁòËáÍ­
ÎÞË®ÁòËáÍ­
£¬ÊµÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷
X·Ö½â²úÉúµÄË®ÕôÆøÈ«²¿±»Å¨ÁòËáÎüÊÕ
X·Ö½â²úÉúµÄË®ÕôÆøÈ«²¿±»Å¨ÁòËáÎüÊÕ
£»
¢ÛÈçºÎÅжÏXÒÑÍêÈ«·Ö½â£¿
Á¬ÐøÁ½´Î¼ÓÈÈ¡¢¹ÄÆø¡¢ÀäÈ´¡¢³ÆÁ¿¼××°ÖõÄÖÊÁ¿£¬
Á¬ÐøÁ½´Î¼ÓÈÈ¡¢¹ÄÆø¡¢ÀäÈ´¡¢³ÆÁ¿¼××°ÖõÄÖÊÁ¿£¬
£»
ÖÊÁ¿²î²»³¬¹ý0.1g
ÖÊÁ¿²î²»³¬¹ý0.1g
£®
¢Ü¸ü¾«È·µÄ²â¶¨µÃ³öÈçÏÂÊý¾Ý£ºXÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ16.52g±äΪ12.00g£¬×°ÖÃÒÒÔöÖØ1.44g£®Ð´³öXµÄ»¯Ñ§Ê½
7CuCO3?8Cu£¨OH£©2»òCu15£¨OH£©16£¨CO3£©7»ò15CuO?7CO2?8H2O
7CuCO3?8Cu£¨OH£©2»òCu15£¨OH£©16£¨CO3£©7»ò15CuO?7CO2?8H2O
£®
Çëд³öXÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º
7CuCO3?8Cu£¨OH£©2+30HCl=15CuCl2+7CO2¡ü+23H2O
7CuCO3?8Cu£¨OH£©2+30HCl=15CuCl2+7CO2¡ü+23H2O
£®
£¨4£©LÊÇÓÉ3ÖÖÔªËع¹³ÉµÄ·Ö×Ó£¬ÄÜÓëIÒÔ1£º2µÄÎïÖʵÄÁ¿Ö®±È·´Ó¦Éú³ÉÄòËØCO£¨NH2£©2ºÍÎïÖÊM£¬ÆäÖÐMÄÜʹÏõËáËữµÄÏõËáÒøÈÜÒº²úÉú°×É«³Áµí£¬Ð´³öLµÄµç×Óʽ
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖªXÊÇÒ»ÖÖÑΣ¬HÊÇÒ»ÖֹŴúÀͶ¯ÈËÃñ¾ÍÒÑÕÆÎÕÒ±Á¶¼¼ÊõµÄ½ðÊôµ¥ÖÊ£¬F¡¢P¡¢JÊdz£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬I¡¢E¡¢G¶¼Êǹ¤ÒµÉÏÖØÒªµÄ¼îÐÔÎïÖÊ£¬PºÍJÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉI¡£ËüÃÇÓÐÓÒͼËùʾµÄ¹Øϵ£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©GµÄ»¯Ñ§Ê½Îª          £»

£¨2£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º

    ¢Ù                     £»         

¢Ú                    £»

£¨3£©Ä³¿ÎÍâС×éͬѧÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨Ä³Ð©×°ÖÃÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬½øÒ»²½Ì½¾¿XÖи÷ÔªËصÄÖÊÁ¿¹Øϵ¡£

    ¢Ù ΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ£»

¢Ú Ïò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ

____________________________________________£»

    ±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆΪ________________£¬ÊµÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷____________________________________________£»

    ¢Û ÈçºÎÅжÏXÒÑÍêÈ«·Ö½â£¿

    __________________________________________________

    ___________________________________________________________________________________________________¡£

    ¢Ü ¸ü¾«È·µÄ²â¶¨µÃ³öÈçÏÂÊý¾Ý£ºXÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ16£®52 g±äΪ12£®00 g£¬×°ÖÃÒÒÔöÖØ1£®44 g¡£Ð´³öXµÄ»¯Ñ§Ê½________________________¡£

    Çëд³öXÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                              ¡¡      

£¨4£©LÊÇÓÉ3ÖÖÔªËع¹³ÉµÄ·Ö×Ó£¬ÄÜÓëIÒÔ1¡Ã2µÄÎïÖʵÄÁ¿Ö®±È·´Ó¦Éú³ÉÄòËØ

CO£¨NH2£©2 ºÍÎïÖÊM£¬ÆäÖÐMÄÜʹÏõËáËữµÄÏõËáÒøÈÜÒº²úÉú°×É«³Áµí£¬Ð´³öLµÄµç×Óʽ                ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìºÓÄÏʡʾ·¶ÐÔ¸ßÖиßÈýÎåУÁªÒêÄ£Ä⿼ÊÔ£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªXÊÇÒ»ÖÖÑΣ¬HÊÇÒ»ÖֹŴúÀͶ¯ÈËÃñ¾ÍÒÑÕÆÎÕÒ±Á¶¼¼ÊõµÄ½ðÊôµ¥ÖÊ£¬F¡¢P¡¢JÊdz£¼ûµÄ·Ç½ðÊôµ¥ÖÊ£¬I¡¢E¡¢G¶¼Êǹ¤ÒµÉÏÖØÒªµÄ¼îÐÔÎïÖÊ£¬PºÍJÔÚÒ»¶¨Ìõ¼þÏ·´Ó¦Éú³ÉI¡£ËüÃÇÓÐÓÒͼËùʾµÄ¹Øϵ£¬ÊԻشðÏÂÁÐÎÊÌ⣺

£¨1£©GµÄ»¯Ñ§Ê½Îª          £»
£¨2£©Ð´³öÏÂÁз´Ó¦µÄ»¯Ñ§·½³Ìʽ£º
¢Ù                     £»         
¢Ú                    £»
£¨3£©Ä³¿ÎÍâС×éͬѧÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨Ä³Ð©×°ÖÃÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬½øÒ»²½Ì½¾¿XÖи÷ÔªËصÄÖÊÁ¿¹Øϵ¡£
¢Ù ΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ£»
¢Ú Ïò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ

____________________________________________£»
±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆΪ________________£¬ÊµÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷____________________________________________£»
¢Û ÈçºÎÅжÏXÒÑÍêÈ«·Ö½â£¿
__________________________________________________
___________________________________________________________________________________________________¡£
¢Ü ¸ü¾«È·µÄ²â¶¨µÃ³öÈçÏÂÊý¾Ý£ºXÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ16£®52 g±äΪ12£®00 g£¬×°ÖÃÒÒÔöÖØ1£®44 g¡£Ð´³öXµÄ»¯Ñ§Ê½________________________¡£
Çëд³öXÓëÑÎËá·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º                              ¡¡      
£¨4£©LÊÇÓÉ3ÖÖÔªËع¹³ÉµÄ·Ö×Ó£¬ÄÜÓëIÒÔ1¡Ã2µÄÎïÖʵÄÁ¿Ö®±È·´Ó¦Éú³ÉÄòËØ
CO£¨NH2£©2 ºÍÎïÖÊM£¬ÆäÖÐMÄÜʹÏõËáËữµÄÏõËáÒøÈÜÒº²úÉú°×É«³Áµí£¬Ð´³öLµÄµç×Óʽ                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìÕã½­Ê¡´ÈϪÖÐѧ¸ßÈýÉÏѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÊµÑéÌâ

£¨6·Ö£©Ä³¿ÎÍâС×éͬѧÉè¼ÆÁËÏÂÁÐʵÑé×°Öã¬Í¨¹ý²â¶¨Ä³Ð©×°ÖÃÖÐÊÔ¼ÁµÄÖÊÁ¿±ä»¯£¬Ì½¾¿Ä³¿×ȸʯ[»¯Ñ§Ê½ÎªaCuCO3¡¤bCu(OH)2£¬a¡¢bΪÕýÕûÊý]Öи÷ÔªËصÄÖÊÁ¿¹Øϵ¡£

¢ÙΪʹÊý¾Ý׼ȷ£¬»¹Ðè²¹³ä×°Öã¬ÇëÄãÔÚ·½¿òÄÚ»æ³ö×°ÖÃͼ²¢Ð´³öÊÔ¼ÁÃû³Æ£»£¨×¢£º¼îʯ»Ò¿É¿´×÷ÉÕ¼îºÍÉúʯ»ÒµÄ»ìºÏÎ
¢ÚÏò×°ÖÃÖйÄÈë¿ÕÆøµÄÄ¿µÄÊÇ__¡¡¡¡¡¡¡¡¡¡________                        £»
±û×°ÖÃÖÐÒ©Æ·µÄÃû³ÆΪ________       £¬
ʵÑéʱ£¬¸ÃҩƷδ¼ûÃ÷ÏԱ仯£¬Ö¤Ã÷____________________________________________£»
¢ÛÁ¬ÐøÁ½´Î¼ÓÈÈ¡¢¹ÄÆø¡¢ÀäÈ´¡¢³ÆÁ¿¼××°ÖõÄÖÊÁ¿£¬ÖÊÁ¿²î²»³¬¹ý0.1 g£¬Åжϸÿ×ȸʯÒÑÍêÈ«·Ö½â¡£
¢Ü¸ü¾«È·µÄ²â¶¨µÃ³öÈçÏÂÊý¾Ý£º¿×ȸʯÊÜÈȺóÍêÈ«·Ö½â£¬¹ÌÌåÓÉ16.52 g±äΪ12.00 g£¬×°ÖÃÒÒÔöÖØ1£®44 g¡£Ð´³ö¸Ã¿×ȸʯµÄ»¯Ñ§Ê½________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸